Categories: Physics, Plasma physics.

# Lawson criterion

For sustained nuclear fusion to be possible,
the **Lawson criterion** must be met,
from which some required properties
of the plasma and the reactor chamber can be deduced.

Suppose that a reactor generates a given power $P_\mathrm{fus}$ by nuclear fusion, but that it leaks energy at a rate $P_\mathrm{loss}$ in an unusable way. If an auxiliary input power $P_\mathrm{aux}$ sustains the fusion reaction, then the following inequality must be satisfied in order to have harvestable energy:

$\begin{aligned} P_\mathrm{loss} \le P_\mathrm{fus} + P_\mathrm{aux} \end{aligned}$We can rewrite $P_\mathrm{aux}$ using the definition
of the **energy gain factor** $Q$,
which is the ratio of the output and input powers of the fusion reaction:

Returning to the inequality, we can thus rearrange its right-hand side as follows:

$\begin{aligned} P_\mathrm{loss} \le P_\mathrm{fus} + \frac{P_\mathrm{fus}}{Q} = P_\mathrm{fus} \Big( 1 + \frac{1}{Q} \Big) = P_\mathrm{fus} \Big( \frac{Q + 1}{Q} \Big) \end{aligned}$We assume that the plasma has equal species densities $n_i = n_e$, so its total density $n = 2 n_i$. Then $P_\mathrm{fus}$ is as follows, where $f_{ii}$ is the frequency with which a given ion collides with other ions, and $E_\mathrm{fus}$ is the energy released by a single fusion reaction:

$\begin{aligned} P_\mathrm{fus} = f_{ii} n_i E_\mathrm{fus} = \big( n_i \Expval{\sigma v} \big) n_i E_\mathrm{fus} = \frac{n^2}{4} \Expval{\sigma v} E_\mathrm{fus} \end{aligned}$Where $\Expval{\sigma v}$ is the mean product of the velocity $v$ and the collision cross-section $\sigma$.

Furthermore, assuming that both species have the same temperature $T_i = T_e = T$, the total energy density $W$ of the plasma is given by:

$\begin{aligned} W = \frac{3}{2} k_B T_i n_i + \frac{3}{2} k_B T_e n_e = 3 k_B T n \end{aligned}$Where $k_B$ is Boltzmann’s constant.
From this, we can define the **confinement time** $\tau_E$
as the characteristic lifetime of energy in the reactor, before leakage.
Therefore:

Inserting these new expressions for $P_\mathrm{fus}$ and $P_\mathrm{loss}$ into the inequality, we arrive at:

$\begin{aligned} \frac{3 n k_B T}{\tau_E} \le \frac{n^2}{4} \Expval{\sigma v} E_\mathrm{fus} \Big( \frac{Q + 1}{Q} \Big) \end{aligned}$This can be rearranged to the form below, which is the original Lawson criterion:

$\begin{aligned} n \tau_E \ge \frac{Q}{Q + 1} \frac{12 k_B T}{\Expval{\sigma v} E_\mathrm{fus}} \end{aligned}$However, it turns out that the highest fusion power density is reached when $T$ is at the minimum of $T^2 / \Expval{\sigma v}$. Therefore, we multiply by $T$ to get the Lawson triple product:

$\begin{aligned} \boxed{ n T \tau_E \ge \frac{Q}{Q + 1} \frac{12 k_B T^2}{\Expval{\sigma v} E_\mathrm{fus}} } \end{aligned}$For some reason, it is often assumed that the fusion is infinitely profitable $Q \to \infty$, in which case the criterion reduces to:

$\begin{aligned} n T \tau_E \ge \frac{12 k_B T^2}{\Expval{\sigma v} E_\mathrm{fus}} \end{aligned}$## References

- M. Salewski, A.H. Nielsen,
*Plasma physics: lecture notes*, 2021, unpublished.